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A gas consisting of diatomic molecules is at a temperature of 117 degree Celsius. If the moment of inertia of each molecule about an axis passing through the centre and normal to the molecule axis is 1.5 ×10^-40 kgm2 find the rms angular speed of each molecule (Kb = 1.38×10^-23Jk^-1) (1)16×10^9 rad per sec (2)12×10^9 rad per sec (3)8×10^9 rad per sec (4)6×10^9 rad per sec?
correct answer is(4) how?
correct answer is(4) how?
Most Upvoted Answer
A gas consisting of diatomic molecules is at a temperature of 117 degr...
Rotational K.E = 2 ×Kb T/2. (as rotational degree of freedom f = 2. for diatomic molecules) & rotational K.E for axis passing through the center is K.E= 2 ×I w^2/2. .... equating both equation we get Kb T= Iw^2.. since, rms angular speed w= sqrt. Kb T/ I. after putting value we get option 4.
Community Answer
A gas consisting of diatomic molecules is at a temperature of 117 degr...
To find the rms (root mean square) angular speed of each molecule, we can use the formula:
ω = √(3kT / I)
where:
ω is the rms angular speed
k is the Boltzmann constant (1.38×10^-23 J/K)
T is the temperature in Kelvin
I is the moment of inertia of each molecule
Given:
Temperature, T = 117 degrees Celsius = 117 + 273 = 390 K
Moment of inertia, I = 1.5 × 10^-40 kgm^2
Boltzmann constant, k = 1.38×10^-23 J/K
Now, let's calculate the rms angular speed using the formula.
1. Convert the temperature to Kelvin:
T = 390 K
2. Plug in the values into the formula:
ω = √(3 * 1.38×10^-23 J/K * 390 K / 1.5 × 10^-40 kgm^2)
3. Simplify the equation:
ω = √(3 * 1.38×10^-23 J/K * 390 K / 1.5 × 10^-40 kgm^2)
= √(3 * 1.38×10^-23 J * 390 / 1.5 × 10^-40 kgm^2)
= √(3 * 1.38×10^-23 J * 390 / 1.5 × 10^-40 kgm^2)
= √(3 * 1.38×10^-23 J * 390 / 1.5 × 10^-40 kgm^2)
= √(3 * 1.38×10^-23 J * 390 / 1.5 × 10^-40 kgm^2)
= √(3 * 1.38×10^-23 J * 390 / 1.5 × 10^-40 kgm^2)
= √(3 * 1.38×10^-23 J * 390 / 1.5 × 10^-40 kgm^2)
= √(3 * 1.38×10^-23 J * 390 / 1.5 × 10^-40 kgm^2)
= √(3 * 1.38×390 / 1.5 × 10^17)
= √(3 * 537.3 / 1.5 × 10^17)
= √(1611.9 / 1.5 × 10^17)
= √(0.0108 × 10^17)
= √(1.08 × 10^15)
= 3.29 × 10^7 rad/s
Therefore, the rms angular speed of each molecule is approximately 3.29 × 10^7 rad/s, which is not one of the given answer options. It seems there might be an error in the given options or the calculation.
ω = √(3kT / I)
where:
ω is the rms angular speed
k is the Boltzmann constant (1.38×10^-23 J/K)
T is the temperature in Kelvin
I is the moment of inertia of each molecule
Given:
Temperature, T = 117 degrees Celsius = 117 + 273 = 390 K
Moment of inertia, I = 1.5 × 10^-40 kgm^2
Boltzmann constant, k = 1.38×10^-23 J/K
Now, let's calculate the rms angular speed using the formula.
1. Convert the temperature to Kelvin:
T = 390 K
2. Plug in the values into the formula:
ω = √(3 * 1.38×10^-23 J/K * 390 K / 1.5 × 10^-40 kgm^2)
3. Simplify the equation:
ω = √(3 * 1.38×10^-23 J/K * 390 K / 1.5 × 10^-40 kgm^2)
= √(3 * 1.38×10^-23 J * 390 / 1.5 × 10^-40 kgm^2)
= √(3 * 1.38×10^-23 J * 390 / 1.5 × 10^-40 kgm^2)
= √(3 * 1.38×10^-23 J * 390 / 1.5 × 10^-40 kgm^2)
= √(3 * 1.38×10^-23 J * 390 / 1.5 × 10^-40 kgm^2)
= √(3 * 1.38×10^-23 J * 390 / 1.5 × 10^-40 kgm^2)
= √(3 * 1.38×10^-23 J * 390 / 1.5 × 10^-40 kgm^2)
= √(3 * 1.38×10^-23 J * 390 / 1.5 × 10^-40 kgm^2)
= √(3 * 1.38×390 / 1.5 × 10^17)
= √(3 * 537.3 / 1.5 × 10^17)
= √(1611.9 / 1.5 × 10^17)
= √(0.0108 × 10^17)
= √(1.08 × 10^15)
= 3.29 × 10^7 rad/s
Therefore, the rms angular speed of each molecule is approximately 3.29 × 10^7 rad/s, which is not one of the given answer options. It seems there might be an error in the given options or the calculation.
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A gas consisting of diatomic molecules is at a temperature of 117 degree Celsius. If the moment of inertia of each molecule about an axis passing through the centre and normal to the molecule axis is 1.5 ×10^-40 kgm2 find the rms angular speed of each molecule (Kb = 1.38×10^-23Jk^-1) (1)16×10^9 rad per sec (2)12×10^9 rad per sec (3)8×10^9 rad per sec (4)6×10^9 rad per sec? correct answer is(4) how?
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A gas consisting of diatomic molecules is at a temperature of 117 degree Celsius. If the moment of inertia of each molecule about an axis passing through the centre and normal to the molecule axis is 1.5 ×10^-40 kgm2 find the rms angular speed of each molecule (Kb = 1.38×10^-23Jk^-1) (1)16×10^9 rad per sec (2)12×10^9 rad per sec (3)8×10^9 rad per sec (4)6×10^9 rad per sec? correct answer is(4) how? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A gas consisting of diatomic molecules is at a temperature of 117 degree Celsius. If the moment of inertia of each molecule about an axis passing through the centre and normal to the molecule axis is 1.5 ×10^-40 kgm2 find the rms angular speed of each molecule (Kb = 1.38×10^-23Jk^-1) (1)16×10^9 rad per sec (2)12×10^9 rad per sec (3)8×10^9 rad per sec (4)6×10^9 rad per sec? correct answer is(4) how? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A gas consisting of diatomic molecules is at a temperature of 117 degree Celsius. If the moment of inertia of each molecule about an axis passing through the centre and normal to the molecule axis is 1.5 ×10^-40 kgm2 find the rms angular speed of each molecule (Kb = 1.38×10^-23Jk^-1) (1)16×10^9 rad per sec (2)12×10^9 rad per sec (3)8×10^9 rad per sec (4)6×10^9 rad per sec? correct answer is(4) how?.
A gas consisting of diatomic molecules is at a temperature of 117 degree Celsius. If the moment of inertia of each molecule about an axis passing through the centre and normal to the molecule axis is 1.5 ×10^-40 kgm2 find the rms angular speed of each molecule (Kb = 1.38×10^-23Jk^-1) (1)16×10^9 rad per sec (2)12×10^9 rad per sec (3)8×10^9 rad per sec (4)6×10^9 rad per sec? correct answer is(4) how? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A gas consisting of diatomic molecules is at a temperature of 117 degree Celsius. If the moment of inertia of each molecule about an axis passing through the centre and normal to the molecule axis is 1.5 ×10^-40 kgm2 find the rms angular speed of each molecule (Kb = 1.38×10^-23Jk^-1) (1)16×10^9 rad per sec (2)12×10^9 rad per sec (3)8×10^9 rad per sec (4)6×10^9 rad per sec? correct answer is(4) how? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A gas consisting of diatomic molecules is at a temperature of 117 degree Celsius. If the moment of inertia of each molecule about an axis passing through the centre and normal to the molecule axis is 1.5 ×10^-40 kgm2 find the rms angular speed of each molecule (Kb = 1.38×10^-23Jk^-1) (1)16×10^9 rad per sec (2)12×10^9 rad per sec (3)8×10^9 rad per sec (4)6×10^9 rad per sec? correct answer is(4) how?.
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